What is the value of the following logarithm? $\log_{13} \left(\dfrac{1}{13}\right)$
Solution: If $b^y = x$ , then $\log_{b} x = y$ Therefore, we want to find the value $y$ such that $13^{y} = \dfrac{1}{13}$ Any number raised to the power $-1$ is its reciprocal, so $13^{-1} = \dfrac{1}{13}$ and thus $\log_{13} \left(\dfrac{1}{13}\right) = -1$.